python - 在 python 中交换列表元素，其中表达式包含函数调用

Question

如果 arr = [4,3,2,1] 并且我想将第一个值与数组的最小值交换，如果我在 python 上使用它的话

arr[0] , arr[arr.index(min(arr))] = min(arr) , arr[0] 
#or   
arr[0] , arr[arr.index(min(arr))] = arr[arr.index(min(arr))] , arr[0]

他们不工作，但如果我这样做

b = arr.index(min(arr))  
#and then  
arr[0] , arr[b] = arr[b] , arr[0]

这很好用。谁能解释为什么？

Answer 1

这与操作的顺序有关。

我们可以对列表进行子类化以对其进行检测以显示它们在做什么。

class TracingList(list):
    def __getitem__(self, key):
        value = super().__getitem__(key)
        print(self, "reading", key, "=", value)
        return value

    def __setitem__(self, key, value):
        print(self, "writing", key, "=", value)
        super().__setitem__(key, value)

    def index(self, value):
        index = super().index(value)
        print(self, "finding index of", value, "=", index)
        return index


arr = TracingList([4, 3, 2, 1])
arr[0], arr[arr.index(min(arr))] = min(arr), arr[0]
print(arr)
print("===")

arr = TracingList([4, 3, 2, 1])
arr[0], arr[arr.index(min(arr))] = arr[arr.index(min(arr))], arr[0]
print(arr)
print("===")

arr = TracingList([4, 3, 2, 1])
b = arr.index(min(arr))
arr[0], arr[b] = arr[b], arr[0]
print(arr)

打印出来

[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] finding index of 1 = 0
[1, 3, 2, 1] writing 0 = 4
[4, 3, 2, 1]
===
[4, 3, 2, 1] finding index of 1 = 3
[4, 3, 2, 1] reading 3 = 1
[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] finding index of 1 = 0
[1, 3, 2, 1] writing 0 = 4
[4, 3, 2, 1]
===
[4, 3, 2, 1] finding index of 1 = 3
[4, 3, 2, 1] reading 3 = 1
[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] writing 3 = 4
[1, 3, 2, 4]